Find the minimum of $xy+yz+zx+\frac1x+\frac2y+\frac5z$ for $x, y, z > 0$

Question

Let $x,y,z > 0$. What is the smallest value of the below expression? \begin{align*} xy+yz+zx+\frac1x+\frac2y+\frac5z \end{align*}

I tried with Lagrange multiplier but I have not any constraint.

Answer 1

There appears to be a local minimum at $(0.605720276464826, 0.908549253319499, 1.817108635900953)$ [numerical search].

Have you tried to apply the criterion for minima of a scalar function of several variables on $\mathbb{R}^3$ and after that applying your restriction that all input values are positive?

The only zero of the gradient in the first quadrant is at $x=\frac{6^{1/3}}{3},y=\frac{6^{1/3}}{2},z=6^{1/3}$ with value of your function at that point equal to $3\,6^{2/3}\approx 9.905781747$.

Of course we still have do additional work to verify this is the solution.

Answer 2

Remarks: While we used calculus to motivate our solution (the details are given at the end), we do not need to include any calculus in the solution!

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Let $a = \frac{x}{z} > 0$ and $b = \frac{y}{z} > 0$ (correspondingly, $x = az, ~ y = bz$).

We have \begin{align*} &xy + yz + zx + \frac{1}{x} + \frac{2}{y} + \frac{5}{z}\\[5pt] =\,& z^2(ab + b + a) + \frac{1}{z}\left(\frac{1}{a} + \frac{2}{b} + 5\right)\\[5pt] =\,& z^2(ab + b + a) + \frac{1}{2z}\left(\frac{1}{a} + \frac{2}{b} + 5\right) + \frac{1}{2z}\left(\frac{1}{a} + \frac{2}{b} + 5\right)\\[5pt] \ge\,& 3\sqrt[3]{z^2(ab + b + a) \cdot \frac{1}{2z}\left(\frac{1}{a} + \frac{2}{b} + 5\right) \cdot \frac{1}{2z}\left(\frac{1}{a} + \frac{2}{b} + 5\right)}\\[5pt] =\,& 3\sqrt[3]{\frac14(ab + b + a)\left(\frac{1}{a} + \frac{2}{b} + 5\right)^2}\\ \ge\,& 3\sqrt[3]{\frac14 \cdot 144}\\ =\,& 3\cdot 6^{2/3} \end{align*} where we have used $$(ab + b + a)\left(\frac{1}{a} + \frac{2}{b} + 5\right)^2 \ge 144. \tag{1}$$ (We give two proofs of (1) at the end.)

Also, when $x = \frac13 6^{1/3}, y = \frac12 6^{1/3}, z = 6^{1/3}$, we have $xy + yz + zx + \frac{1}{x} + \frac{2}{y} + \frac{5}{z} = 3 \cdot 6^{2/3}$.

Thus, the minimum of $xy + yz + zx + \frac{1}{x} + \frac{2}{y} + \frac{5}{z}$ is $3\cdot 6^{2/3}$.

We are done.

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$\phantom{2}$

The first proof of (1):

Letting $a = u/3, b = v/2$, it suffices to prove that \begin{align*} &25\,{u}^{3}{v}^{3}+90\,{u}^{3}{v}^{2}+105\,{u}^{2}{v}^{3}+96\,{u}^{3}v +99\,u{v}^{3}\\ &\quad +32\,{u}^{3}+96\,{u}^{2}v+90\,u{v}^{2 }+27\,{v}^{3} \ge 660\,{u}^{2}{v}^{2}. \end{align*} Using AM-GM, we have \begin{align*} &\mathrm{LHS}\\ \ge\,& 660\sqrt[660]{(u^3v^3)^{25} (u^3v^2)^{90} (u^2v^3)^{105} (u^3v)^{96} (uv^3)^{99} (u^3)^{32} (u^2v)^{96} (uv^2)^{90} (v^3)^{27}}\\ =\,& 660u^2v^2. \end{align*} (Note: $25+90+105+96+99+32+96+90+27 = 660$.)

We are done.

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The second proof of (1):

Let $u = ab + b + a$. Using Cauchy-Bunyakovsky-Schwarz inequality, we have $$ \frac{1}{a} + \frac{2}{b} + 5 = \frac{1}{a} + \frac{2 + 2u}{u - a} + 3 \ge \frac{(1 + \sqrt{2+2u})^2}{a + u - a} + 3 = \frac{(1 + \sqrt{2+2u})^2}{ u } + 3. $$ Thus, $$(ab + b + a)\left(\frac{1}{a} + \frac{2}{b} + 5\right)^2 \ge u\left(\frac{(1 + \sqrt{2+2u})^2}{u} + 3\right)^2.$$

It suffices to prove that, for all $u > 0$, $$u\left(\frac{(1 + \sqrt{2+2u})^2}{u} + 3\right)^2 \ge 144.$$

Letting $2 + 2u = v^2$ for $v > \sqrt 2$, we have $$u\left(\frac{(1 + \sqrt{2+2u})^2}{u} + 3\right)^2 - 144 = \frac{(25v^2 + 140v + 148)(v - 2)^2}{2(v^2 - 2)} \ge 0.$$

We are done.

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$\phantom{2}$

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Details of using calculus to motivate the solution:

(1) Let $x = az, y = bz$. We have $$xy + yz + zx + \frac{1}{x} + \frac{2}{y} + \frac{5}{z} = z^2(ab + b + a) + \frac{1}{z}\left(\frac{1}{a} + \frac{2}{b} + 5\right).$$ We can appy AM-GM to get $$xy + yz + zx + \frac{1}{x} + \frac{2}{y} + \frac{5}{z} \ge 3\sqrt[3]{\frac14(ab + b + a)\left(\frac{1}{a} + \frac{2}{b} + 5\right)^2}.$$

We need to find the minimum of $$f(a, b) := (ab + b + a)\left(\frac{1}{a} + \frac{2}{b} + 5\right)^2, \quad a, b > 0.$$ We have \begin{align*} \frac{\partial f}{\partial a} &= {\frac { \left( 5\,ab+2\,a+b \right) \left( 5\,{a}^{2}{b}^{2}+7\,{a}^ {2}b-a{b}^{2}+2\,{a}^{2}-ab-2\,{b}^{2} \right) }{{a}^{3}{b}^{2}}},\\ \frac{\partial f}{\partial b} &= {\frac { \left( 5\,ab+2\,a+b \right) \left( 5\,{a}^{2}{b}^{2}-2\,{a}^ {2}b+6\,a{b}^{2}-4\,{a}^{2}-2\,ab+{b}^{2} \right) }{{a}^{2}{b}^{3}}}. \end{align*} From $\frac{\partial f}{\partial a} = \frac{\partial f}{\partial b} = 0$, we have \begin{align*} 5\,{a}^{2}{b}^{2}+7\,{a}^{2}b-a{b}^{2}+2\,{a}^{2}-ab-2\,{b}^{2} &= 0, \tag{2}\\ 5\,{a}^{2}{b}^{2}-2\,{a}^{2}b+6\,a{b}^{2}-4\,{a}^{2}-2\,ab+{b}^{2} &= 0. \tag{3} \end{align*} $(2) \times (5b^2-2b-4) - (3) \times (5b^2+7b+2)$ gives $$-b \left( 35\,{b}^{3}a+35\,a{b}^{2}+15\,{b}^{3}-8\,ab+3\,{b}^{2}-8\,a- 6\,b \right) = 0$$ which results in $$a = -3\,{\frac {b \left( 5\,{b}^{2}+b-2 \right) }{35\,{b}^{3}+35\,{b}^{2}- 8\,b-8}}.$$ Then (1) becomes $$-4\,{\frac {{b}^{2} \left( 5\,b+2 \right) \left( 4\,b+1 \right) \left( 2\,b-1 \right) \left( 5\,{b}^{2}+5\,b-1 \right) }{ \left( 35 \,{b}^{2}-8 \right) ^{2} \left( b+1 \right) }} = 0$$ which results in $b = 1/2$. Then $a = 1/3$.

(2) Early attempt:

We have \begin{align*} xy + yz + zx + \frac{1}{x} + \frac{2}{y} + \frac{5}{z} &= (y + z)x + \frac{1}{x} + \frac{2}{y} + yz + \frac{5}{z}\\ &\ge 2\sqrt{y + z} + \frac{2}{y} + yz + \frac{5}{z}. \end{align*} We need to find the minimum of $$g(y, z) := 2\sqrt{y + z} + \frac{2}{y} + yz + \frac{5}{z}, \quad y, z > 0.$$ We have \begin{align*} \frac{\partial g}{\partial y} &= \frac{1}{\sqrt{y+z}} - \frac{2}{y^2} + z, \\ \frac{\partial g}{\partial z} &= \frac{1}{\sqrt{y+z}} + y - \frac{5}{z^2}. \end{align*} After some complicated manipulations, we get $y = \frac12 6^{1/3}$ and $z = 6^{1/3}$. If computer is used, the reasoning is simple.